What is the solution for 3xyy^'+4x-2=0 ?

The solution for 3xyy^'+4x-2=0 is y=sqrt((2(-4x+2ln(x))+c_{1))/3},y=-sqrt((2(-4x+2ln(x))+c_{1))/3}

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3xyy^'+4x-2=0

3 x y y^{^{'}} + 4 x - 2 = 0

y = \frac{2 ( - 4 x + 2 ln ( x ) ) + c _{1}}{3}, y = - \frac{2 ( - 4 x + 2 ln ( x ) ) + c _{1}}{3}

Solution steps

3 x y y^{'} (x) + 4 x - 2 = 0

Solve separable ODE:

y = \frac{2 ( - 4 x + 2 ln ( x ) ) + c _{1}}{3}, y = - \frac{2 ( - 4 x + 2 ln ( x ) ) + c _{1}}{3}

y = \frac{2 ( - 4 x + 2 ln ( x ) ) + c _{1}}{3}, y = - \frac{2 ( - 4 x + 2 ln ( x ) ) + c _{1}}{3}

\frac{d y}{d x} = \frac{( 3 x + x y ^{2} )}{( 2 y - x ^{2} y )}

x (1 + y^{2})^{\frac{1}{2}} d x = y d y

\frac{d y}{d x} = y^{2}, y (0) = 1

\frac{d y}{d x} - x^{3} y^{2} = y^{2}

\frac{x d y}{d x} - y = 7 x^{6} y

What is the solution for 3xyy^'+4x-2=0 ?
The solution for 3xyy^'+4x-2=0 is y=sqrt((2(-4x+2ln(x))+c_{1))/3},y=-sqrt((2(-4x+2ln(x))+c_{1))/3}